A 1.00 Kg ball is shot out of a cannon inclined at 40 degrees, by a spring initi

Question

A 1.00 Kg ball is shot out of a cannon inclined at 40 degrees, by a spring initially compressed by 0.500 m. The strength constant of the spring is 10,000 N/m; The coefficient of kinetic frictions between the ball and the barrel of the cannon is .1; the initial distance from the ball to the end of the barrel is 8.5 m; and the barrel is 9.00m long.

a) What is the closed system ? (neglect air resistance)

b) What is the velocity of the ball at the apex of its motion?

c) Where does the ball land relative to the cannon?

Explanation / Answer

a) a system in which there is no loss of energy to the sorroundings is called closed system

b)

initial energy

Ei = 0.5 k x^2 = 0.5 * 10000 * 0.5*0.5 = 1.250 k J

final energy

Ef = Ei - m g d sin40 - u m g d cos40 = 1250 - 1*9.8*9*sin*40 - 0.1*1*9.8*9*cos40 = 1.18655 kJ

Ef = 0.5 m v^2

1186.5 = 0.5 * 1 * v^2

v = 48.714 m/s

vx = v cos40 = 48.714 * cos40 = 37.32 m/s

c) xo = 9*cos40 = 6.89 m

R = V^2*sin2*40/g =48.71^2*sin80 / 9.8 = 238.43m

distance from barallel = xo+R = 245.32m

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